Question
A person's savings for 3 years are in an arithmetic progression that in the 3 years he has saved cs 24,000 and in the first year he saves more than he saved in the second year
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Answer to a math question A person's savings for 3 years are in an arithmetic progression that in the 3 years he has saved cs 24,000 and in the first year he saves more than he saved in the second year
Murray
4.589 Answers
Let's denote the amount saved in the first year as a d 24,000
1) Amount saved in the 1st year:a
2) Amount saved in the 2nd year:a + d
3) Amount saved in the 3rd year:a + 2d
From the given information, we have the total amount saved in 3 years:
a + (a + d) + (a + 2d) = 24,000
Simplify the equation:
3a + 3d = 24,000
Divide by 3:
a + d = 8,000
Since the amount saved in the 1st year is more than the 2nd year, we have:
a > a + d
Simplify this inequality:
a > 8,000
Therefore, the person's savings for the 3 years area + (a+d) + (a+2d) = 2a + 3d = a + a + 8,000 = 2a + 8,000 8,000
\textbf{Answer:} The person's savings for 3 years are2a + 8,000 a > 8,000
1) Amount saved in the 1st year:
2) Amount saved in the 2nd year:
3) Amount saved in the 3rd year:
From the given information, we have the total amount saved in 3 years:
Simplify the equation:
Divide by 3:
Since the amount saved in the 1st year is more than the 2nd year, we have:
Simplify this inequality:
Therefore, the person's savings for the 3 years are
\textbf{Answer:} The person's savings for 3 years are
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